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astra-53 [7]
3 years ago
6

Heather runs round a circular track many times each day.

Mathematics
1 answer:
slamgirl [31]3 years ago
6 0

Answer:

See explanation

Step-by-step explanation:

1. Determine the total distance heather has run in Km.

4.18km PER DAY for 197 days;  4.18 * 197 =  823.46km.

2. Can we just stop and say that this Heather chick MIGHT HAVE A PROBLEM?! Like, fr, she has run over 500 miles in half a year. . .she needs to see someone about that.

3. Divide the total distance by the distance of the track.

823.46/0.21 = 3921.23809524

4. Conclusion: Heather has run (roughly) 3,931.24 times around the track after the 197 day period of running 4.18km per day.

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How does the indian reservation have the effect on the Right of Occupancy
SVEN [57.7K]

Answer:

As U.S. citizens, American Indians and Alaska Natives are generally subject to federal, state, and local laws. On federal Indian reservations, however, only federal and tribal laws apply to members of the tribe, unless Congress provides otherwise.

Step-by-step explanation:

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3 0
2 years ago
Right triangles abc and dbc with right angle c are given below. If cos(a)=15,ab=12 and cd=2, find the length of bd.
dmitriy555 [2]

see the attached figure to better understand the problem

we have that

cos(A)=\frac{1}{5} \\ AB=12\ units\\ CD=2\ units

Step 1

<u>Find the value of AC</u>

we know that

in the right triangle ABC

cos (A)=(AC/AB)\\AC=AB*cos(A)

substitute the values in the formula

AC=12*(1/5)\\ AC=2.4\ units

Step 2

<u>Find the value of BC</u>

we know that

in the right triangle ABC

Applying the Pythagorean Theorem

AB^{2} =AC^{2}+BC^{2}\\ BC^{2}=AB^{2} -AC^{2}

substitute the values

BC^{2}=12^{2} -2.4^{2}\\BC^{2}= 138.24\\ BC=11.76\ units

Step 3  

<u>Find the value of BD</u>

we know that

in the right triangle BCD

Applying the Pythagorean Theorem

BD^{2} =DC^{2}+BC^{2}

substitute the values  

BD^{2} =2^{2}+11.76^{2}

BD=11.93\ units

therefore

<u>the answer is</u>

the length of BD is 11.93 units

8 0
3 years ago
Read 2 more answers
Use the rule (x,y) (3x,2y) to find the image for the preimage defined by the given points.
Elina [12.6K]

Answer: The points of the images are  (9,10), (15,6), (6,4)  and the image is not a rigid motion because the shape changes in side.

Step-by-step explanation:

Since it gives you the scale factor then find they coordinates by multiplying the coordinates by the scare factor.

A(3,5) → (3*3,5*2) → (9,10)

B( 5,3) → (5*3, 3*2) → (15,6)

C ( 2,3)→ (2*3, 2*2)→ ( 6,4)

4 0
3 years ago
Find the following of each arc measure
Dimas [21]
ZW= 108
VW= 55
UY= 34
8 0
3 years ago
15. If x=a Sin2t (I+Cos2t) and y=b Cos 2t (1-Cos2t) then find<br>dy/dx at =22/7*4<br>​
Sav [38]

By the chain rule,

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm dx}\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}

It looks like we're given

\begin{cases}x=a\sin(2t)(1+\cos(2t))\\y=b\cos(2t)(1-\cos(2t))\end{cases}

where <em>a</em> and <em>b</em> are presumably constant.

Recall that

\cos^2t=\dfrac{1+\cos(2t)}2

\sin^2t=\dfrac{1-\cos(2t)}2

so that

\begin{cases}x=2a\sin(2t)\cos^2t\\y=2b\cos(2t)\sin^2t\end{cases}

Then we have

\dfrac{\mathrm dx}{\mathrm dt}=4a\cos(2t)\cos^2t-4a\sin(2t)\cos t\sin t

\dfrac{\mathrm dy}{\mathrm dt}=-4b\sin(2t)\sin^2t+4b\cos(2t)\sin t\cos t

\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{4b\cos(2t)\sin t\cos t-4b\sin(2t)\sin^2}{4a\cos(2t)\cos^2t-4a\sin(2t)\cos t\sin t}

\implies\boxed{\dfrac{\mathrm dy}{\mathrm dx}=\dfrac ba\tan t}

where the last reduction follows from dividing through everything by \cos(2t)\cos^2t and simplifying.

I'm not sure at which point you're supposed to evaluate the derivative (22/7*4, as in 88/7? or something else?), so I'll leave that to you.

8 0
3 years ago
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